#
Partial-tree procedure
^{
}

The following procedure
`list->tree`

converts an ordered list to a balanced binary tree. The helper procedure
`partial-tree`

takes as arguments an integer
`n`

and list of at least
`n`

elements and constructs a balanced tree containing the first
`n`

elements of the list. The result returned by
`partial-tree`

is a pair (formed with
`cons`

) whose
`car`

is the constructed tree and whose
`cdr`

is the list of elements not included in the tree.

```
(define (list->tree elements)
(car (partial-tree elements (length elements))))
(define (partial-tree elts n)
(if (= n 0)
(cons '() elts)
(let ((left-size (quotient (- n 1) 2)))
(let ((left-result (partial-tree elts left-size)))
(let ((left-tree (car left-result))
(non-left-elts (cdr left-result))
(right-size (- n (+ left-size 1))))
(let ((this-entry (car non-left-elts))
(right-result (partial-tree (cdr non-left-elts)
right-size)))
(let ((right-tree (car right-result))
(remaining-elts (cdr right-result)))
(cons (make-tree this-entry left-tree right-tree)
remaining-elts))))))))
```

a. Write a short paragraph explaining as clearly as you can how
`partial-tree`

works. Draw the tree produced by
`list->tree`

for the list
`(1 3 5 7 9 11)`

b. What is the order of growth in the number of steps required by
`list->tree`

to convert a list of
`n`

elements?